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Backwards Euler method

Posted on 2007-11-19
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The final part of a question presents the "Backwards Euler method", as:

x(n+1) = x(n) + h * f( (n+1) * h,  x(n+1) )              (where f(t,x) = dx/dt)

It then says to apply this method to the linear equation dx/dt=x, and show that the method converges to the true solution x(t)=e^t as t->infinity.

It is obvious that the term (1+t/n)^n will turn up at some stage (seeing as the limit of it, as t->infinity, is e^t). But I'm struggling to get started. My problem is that in order to find the derivative at x(n+1), I must find what x(n+1) is; but to find x(n+1), I need to know the derivative at x(n+1)...

So, can someone at least get me past this hurdle, then I will hopefully be able to figure the rest for myself.

An earlier part of this question might be relevent; it says to apply Euler's method to dx/dt=kx. For which I get:

x(n)=x0(1+hk)^n

...

Thanks
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Question by:Beta07
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by:
ozo earned 2000 total points
ID: 20316803
isn't dx/dt=x the same as  dx/dt=kx with k=1
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by:Beta07
ID: 20319315
Indeed. I suppose you're suggesting that I use:

x(n)=x0(1+hk)^n

(with k=1), to find x(n+1)? I thought about this, but (although it's probably the right way to go), it doesn't really make much sense to me:
Surely by incorporating the standard Euler method into the backwards one, you'd lose some of the advantages of using the backwards method in the first place?

Either way, I'll assume it's the correct method. Which gives me:

x(1) = x(0) + h*x(0)(1+h)
x(2) = x(1) + h*x(0)(1+h)²
       = x(0) + h*x(0)(1+h) + h*x(0)(1+h)²
x(3) = x(2) + h*x(0)(1+h)³
       = x(0) + h*x(0)(1+h) + h*x(0)(1+h)^2 + h*x(0)(1+h)³
...
x(n+1) = x(0){1+h(1+h) + h(1+h)² + h(1+h)³ + ... + h(1+h)^n}
           = x(0){(1+h)^(n+1) - h)}

(Using the equation for the partial sum of a geometric series (and rearranging), in that last step).

Now, this looks quite promising:

x(n+1) = x(0){(1+h)^(n+1) - h)}
=>
x(n) = x(0){(1+h)^n - h}

The question says to use the substitution: h=t/n; thus

x(n) = x(0){(1+t/n)^n - h}

So we're almost there:  the limit as h->0 (or n->infinity)* of
{(1+t/n)^n - h}
is:
e^t

However, I'm not sure how to properly finish this question off. I don't think it would make much sense to say:

lim[h->0]  x(n)/x(0)  =  e^t

So, I guess I'm looking for a rearrangement of:

x(n) = x(0){(1+t/n)^n - h}

such that when I take the limit of both sides, I get x(t) on one side, and e^t on the other....?


Thank you

* In my original question (and also on the question sheet), it says "as t->infinity (equivalently, h->0)"; I can only assume they mean "n->infinity,.. h->0".
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by:Beta07
ID: 20319318
(Please ignore the  characters)
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by:Beta07
ID: 20319330
>>
So, I guess I'm looking for a rearrangement of:

x(n) = x(0){(1+t/n)^n - h}

such that when I take the limit of both sides, I get x(t) on one side, and e^t on the other....?
>>

Or, I suspect, a rearrangement such that when I take the limit of both sides, I get dx/dt on one side, and e^t on the other..?
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Author Comment

by:Beta07
ID: 20320013
Sorry,

> x(n+1) = x(0){1+h(1+h) + h(1+h)² + h(1+h)³ + ... + h(1+h)^n}
              = x(0){(1+h)^(n+1) - h)}

Should be

x(n) = x(0){1+h(1+h) + h(1+h)² + h(1+h)³ + ... + h(1+h)^n}
       = x(0){(1+h)^(n+1) - h)}

Thus

>>
x(n+1) = x(0){(1+h)^(n+1) - h)}
=>
x(n) = x(0){(1+h)^n - h}
>>

Is of course wrong.

So we've got:

x(n) = x(0){(1+h)^(n+1) - h)}
0

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